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(B) Given the equation $\operatorname{Tan}^{-1} x + \operatorname{Tan}^{-1} 2x = \frac{\pi}{4}$.Using the formula $\operatorname{Tan}^{-1} A + \operat

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$1$

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(B) Given the equation $\operatorname{Tan}^{-1} x + \operatorname{Tan}^{-1} 2x = \frac{\pi}{4}$.Using the formula $\operatorname{Tan}^{-1} A + \operatorname{Tan}^{-1} B = \operatorname{Tan}^{-1} \left( \frac{A+B}{1-AB} ight)$,we get:$\operatorname{Tan}^{-1} \left( \frac{x+2x}{1-x(2x)} ight) = \frac{\pi}{4}$.Taking $	an$ on both sides:$\frac{3x}{1-2x^2} = 	an \left( \frac{\pi}{4} ight) = 1$.This simplifies to $3x = 1 - 2x^2$,or $2x^2 + 3x - 1 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm \sqrt{17}}{4}$.Since $\operatorname{Tan}^{-1} x + \operatorname{Tan}^{-1} 2x = \frac{\pi}{4}$ and $\frac{\pi}{4} > 0$,we must have $x > 0$ (as $x$ and $2x$ have the same sign).Checking the values: $x = \frac{-3 + \sqrt{17}}{4} \approx \frac{-3 + 4.12}{4} = 0.28 > 0$ (Valid).$x = \frac{-3 - \sqrt{17}}{4} Thus,there is exactly $1$ real solution.

The number of real solutions of $\operatorname{Tan}^{-1} x + \operatorname{Tan}^{-1} 2x = \frac{\pi}{4}$ is

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